You have two jars and 100 marbles. Fifty of the

marbles are red, and 50 are blue. One of the jars will

be chosen at random, then 1 marble will be withdrawn

from that jar at random. How do you maximize the chance

that a red marble will be chosen? (You must place all 100

marbles in the jars.)What is the chance of selecting

a red marble when using your scheme?

## 13 comments:

1.To maximise the chance of selecting red marbles..put the blue marbles first in the jar..and ask another person to select for the event...

2.Chance of selecting a red marble in this scheme...there r 2 subevents...chance in both the cases is 1/2 considering ideal condition..as both are dependent on one another...total chance of selecting red marble is 1 in 4...

I don't agree with your solution. We have 2 jars and not 1. total 100 marbles out of which 50 are red and 50 are blue coloured. Probability of selecting any one jar is 1/2. Now we want to place these 100 marbles in 2 jars to maximize the probability of red marble pick. We can put 1 red marble in one jar and rest all the 99 marbles those include 50 blue coloured and 49 red coloured in other jar. Now, calculating the probability for red marble pick in the first chance goes this way:

(1/2) * (1/1) = 1/2 considering the first jar in which there is only 1 marble and that is red. If the second marble is considered then it's: (1/2) * (49/99) = 49/198.

So, for calculating the total probability considering both the jars can be done in following way:

((1/2) * (1/1)) + ((1/2) * (49/99)) = (1/2) + (49/198) = 148/198 = 0.7474

This is 74.74 %.

Where as according to your solution the probability of red pick is only 25% (1/4) as you are putting all the 100 marbles in 1 jar only.

I believe I conveyed my scheme clearly. Still, please let me know if there is any query or suggestion or mistake you find. Thank You.

"God" is wrong yet again! If you wish to maximise the chances of selecting a red marble and your scheme gives 25% for red, that must mean that the chance of selecting a blue is 75%! So why wouldn't you swap them around so that your chances of selecting red are 75%?

Señor Dost is correct.

I think one jar should have 1 white marble and the other jar should have 49 white marbles along with 50 black marbles

Where did you get the black and white marbles from?

The probability of red being chosen is ALWAYS the same. The probability only changes after the the jar has been chosen, but by then it's too late. Choosing between the jars is always 50/50. If you stacked one the jars with all red and the other with all blue, then the chances would be:

1/2 * 100% + 1/2* 0 = 50%

It will always be 1/2 x (R+B) which will always be 50% since R + B always ends up being the total of 100

@Adnan - what if you only put 1 red marble in the first jar and all of the rest (49 red + 50 blue) in the second?

Half the time (P=50%) the first jar will be picked, from which ONLY (100%) a red marble can be drawn. (P=50*100/100 = 50%)

Half the time (P=50%) the second jar will be picked, from which there is a 49/(49+50) (=49.49%) chance of drawing a red marble. (P= 50*49.49/100 = 24.74%)

Therefore the total probability of drawing a red marble is 50 + 24.74 = 74.74%

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To maximize the chances put all the red marbles in jar 1 and all blues in other 1. This will make the chances 50-50.

If you can put one red marble in a jar then it follows you are in control of what color and how many marbles go into either jar. Therefore put all the blue marbles in the jars first. Then put the red marbles on top.

Odds of picking a red marble=100%

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